Re: question
- From: Peter Pein <petsie@xxxxxxxxxx>
- Date: Sun, 22 Apr 2007 02:06:06 +0200
David W. Cantrell schrieb:
....
With t -> Pi/4, your 2*EllipticF[Pi/4, 2] becomes 1/2 Beta[1, 1/2, 1/4] and
then version 5.2 gives
In[2]:=
1/2 Beta[1, 1/2, 1/4]
Out[2]=
(Sqrt[Pi]*Gamma[1/4])/(2*Gamma[3/4])
In[3]:=
FullSimplify[% - Gamma[1/4]^2/(2*Sqrt[2*Pi])]
Out[3]=
0
David
Hi,
It is 2AM in germany and I'm far too tired to have a closer look, whether the
identity 1/2 Beta[1, 1/2, 1/4] == Beta[Tan[Pi/8]^2, 1/4, 1/2] is trivial or
not. But I think, it's interesting, that Mathematica gives different forms of
the same expression, when using slightly different substitutions in Dimitri's
integral, because I would have guessed, there is some kind of normalization
inside Mathematica's Integrate[].
ChangeVar[x == Tan[u/4], Integrate[1/Sqrt[Sin[u]], {u, 0, Pi/2}], x]
--> Beta[Tan[Pi/8]^2, 1/4, 1/2]
good night,
Peter
.
- Follow-Ups:
- Re: question
- From: dimitris
- Re: question
- References:
- question
- From: dimitris
- Re: question
- From: Daniel Lichtblau
- Re: question
- From: David W . Cantrell
- question
- Prev by Date: Re: question
- Next by Date: cos(12 degress)*cos(24 degrees)*cos(48 degrees)*cos(96 degrees)
- Previous by thread: Re: question
- Next by thread: Re: question
- Index(es):
Relevant Pages
|
|
